Two cards are drawn from a deck of 52 cards with replacement. What is the probability that: a). Both cards are Queens? b). Both cards are the same suit? c) Both cards have the same face value.<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n Let’s solve each part step by step. Since the two cards are drawn with replacement<\/strong>, the deck is restored<\/strong> after the first draw, so probabilities for the second draw remain the same as the first.<\/p>\n\n\n\n There are 4 Queens<\/strong> in a deck of 52 cards.<\/p>\n\n\n\n There are 4 suits: Spades, Hearts, Diamonds, Clubs \u2014 each with 13 cards.<\/p>\n\n\n\n So, P(both same suit)=1\u00d714=14P(\\text{both same suit}) = 1 \\times \\frac{1}{4} = \\frac{1}{4}<\/p>\n\n\n\n There are 13 different face values (A, 2, 3, …, 10, J, Q, K), and each has 4 cards.<\/p>\n\n\n\n Probability that the second card matches the first’s face value:<\/p>\n\n\n\n So, P(same face value)=113P(\\text{same face value}) = \\frac{1}{13}<\/p>\n\n\n\n a) 1169\\frac{1}{169} In probability, when events occur with replacement<\/strong>, each trial is independent, and the probability distribution remains unchanged. A standard deck has 52 cards, divided into 4 suits and 13 face values.<\/p>\n\n\n\n (a) Both cards are Queens<\/strong>: There are only 4 Queens in the deck. The probability of drawing one Queen is 452\\frac{4}{52}. Since the card is replaced, the same 4 Queens are available for the second draw. Therefore, the events are independent and multiply: 452\u00d7452=1169\\frac{4}{52} \\times \\frac{4}{52} = \\frac{1}{169}. This is quite rare since Queens form a small fraction of the deck.<\/p>\n\n\n\n (b) Both cards are the same suit<\/strong>: Each suit has 13 cards. Drawing any card first doesn’t affect the outcome, but to match the suit in the second draw, only 13 of the 52 cards match. So the probability is 1352=14\\frac{13}{52} = \\frac{1}{4}. This is higher than drawing two specific cards like Queens.<\/p>\n\n\n\n (c) Both cards have the same face value<\/strong>: Each face value (e.g., 7s or Kings) occurs 4 times in the deck. Drawing any card first gives one face value. Then, in the second draw, 4 out of 52 cards match that face value (since it’s replaced). So the chance of drawing the same value again is 452=113\\frac{4}{52} = \\frac{1}{13}, regardless of the specific value.<\/p>\n\n\n\n Understanding independence and how many favorable outcomes exist is key to solving these problems accurately.<\/p>\n","protected":false},"excerpt":{"rendered":" Two cards are drawn from a deck of 52 cards with replacement. What is the probability that: a). Both cards are Queens? b). Both cards are the same suit? c) Both cards have the same face value. The correct answer and explanation is: Let’s solve each part step by step. Since the two cards are […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-28605","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28605","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=28605"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28605\/revisions"}],"predecessor-version":[{"id":28612,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28605\/revisions\/28612"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28605"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28605"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\na) Probability that both cards are Queens<\/strong><\/h3>\n\n\n\n
\n
P(Q1)=452=113P(Q_1) = \\frac{4}{52} = \\frac{1}{13}<\/li>\n\n\n\n
P(Q2)=452=113P(Q_2) = \\frac{4}{52} = \\frac{1}{13}<\/li>\n\n\n\n
P(both\u00a0Queens)=113\u00d7113=1169P(\\text{both Queens}) = \\frac{1}{13} \\times \\frac{1}{13} = \\frac{1}{169}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nb) Probability that both cards are of the same suit<\/strong><\/h3>\n\n\n\n
\n
P(same\u00a0suit)=1352=14P(\\text{same suit}) = \\frac{13}{52} = \\frac{1}{4}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nc) Probability that both cards have the same face value<\/strong><\/h3>\n\n\n\n
\n
\n
P(match)=452=113P(\\text{match}) = \\frac{4}{52} = \\frac{1}{13}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nSummary of Answers:<\/h3>\n\n\n\n
b) 14\\frac{1}{4}
c) 113\\frac{1}{13}<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n